Need Help!- Triggering PC817 with 24VDC, 2AMPS?

[Sunmixed]

Hello guys, I hope i'm in the right place to post this help about triggering PC817 with 24VDC, 2AMPs, for an input pin in parallel port for Mach 3.

Back 4 years ago, i started using and have been using 24VDC Relays to assist me in triggering input pins. The circuit was:



This worked very, very well. However, its very bulky in size. Besides, electronics are vital these days. Wouldn't want to be left out

So, I came up with this schematic:


I soldered it into a PCB Breadboard, all soldering correct, and all of the PCB lines have been enhanced with thicker solder.

Very weird reason of why it doesn't work.

CON1 Pin 1 is fed with 24VDC 2AMPS -
CON1 Pin 2 is fed with 24VDC 2AMPS +

Power is a switching power supply with +/- 0.1V Tolerance.

At the end of the power, i measured 6.9VDC. Looks promising to trigger PC817, however, it doesn't work.

Anyone know why? I'm sorry that it could sound newbie, but this is what i'm struggling at

Thank you in advance!

Sunmixed

[H.O]

Hi,
What's the purpose of the zenerdiode? Try removing it.

[Sunmixed]

Henrik,

I'm not quite sure. I saw on some PCB board who is doing the same thing. @@ Well, I'm gonna remove it soon.

[Torchhead]

That part number is a transit voltage protector (kinda like a zener but has a softer "knee") and will withstand greater overloads. If it's a 6.8 volt device as the part indicates it will grab all the current intended for the opto and not enough will flow to turn it on. The resistor value is right for 24 V (about 10ma) and the cap help reduce any noise from the outside. The TVS diode can go away unless you think the input voltage could spik above about 75VDC. The value needs to be about a 48VDC device. It just protects the opto.

The resistor has to drop about 22.8 V at so the series current is about 9.5 ma. At that the resistor dissipation is 200mw (1/4W) which will run a standard 1/4W resistor on the warm side. Depending on the opto you use you made need a tad more current in the emitter side. Dropping the resistor to a 2.2K would give you wider margin of operation.

The Gain of the opto (stated in the spec sheets) will predict how much current you can get on the other side (NPN transistor). You want to try and run the transistor at fairly close to the suggested current to get decent On and (especially) off times. In most inputs it does not matter and if it worked with relays it certainly will work with a slow opto (:-)

TOM Caudle
www.CandCNC.com
Totally Modular CNC Electronics

[Sunmixed]

Originally Posted by Torchhead That part number is a transit voltage protector (kinda like a zener but has a softer "knee") and will withstand greater overloads. If it's a 6.8 volt device as the part indicates it will grab all the current intended for the opto and not enough will flow to turn it on. The resistor value is right for 24 V (about 10ma) and the cap help reduce any noise from the outside. The TVS diode can go away unless you think the input voltage could spik above about 75VDC. The value needs to be about a 48VDC device. It just protects the opto. The resistor has to drop about 22.8 V at so the series current is about 9.5 ma. At that the resistor dissipation is 200mw (1/4W) which will run a standard 1/4W resistor on the warm side. Depending on the opto you use you made need a tad more current in the emitter side. Dropping the resistor to a 2.2K would give you wider margin of operation. The Gain of the opto (stated in the spec sheets) will predict how much current you can get on the other side (NPN transistor). You want to try and run the transistor at fairly close to the suggested current to get decent On and (especially) off times. In most inputs it does not matter and if it worked with relays it certainly will work with a slow opto (:-) TOM Caudle www.CandCNC.com Totally Modular CNC Electronics

Thank you Tom! Well both of you, I did try removing the 1.5KE6.8, it came out with 23.98VDC.

I suppose i will put it back, and replace it with a 2K2 resistor!

I was wondering, if this 24VDC 2A can be step down into 5V with just this schematic, is it good to power a 5V Relay?

And uhm, can it replace a 7805 and 7812? Just being curious

[H.O]

Hi,
If you have 24V across pin 1 and pin 2 of the PC817 then you either have the pins reversed or the LED in the opto-isolator is shot.

LEDs are current driven and the have a nominal forward voltage drop. In the case of the PC817 this voltage drop is 1.2V typical and the forward current should be somewhere between 10 and 50mA depending on how much current you need to pull thru the transistor on the other side.

If you want to drive the PC817 with a 24V signal, then 22.8V needs to be dropped across the series resistor. Let's say you want 15mA thru the LED then Ohms law (R=U/I) tells you the resistor needs to be 22.8V/0.015=1520ohm. If you want to drive the same PC817 with a 5V signal then resistor needs to be 3.8/0.015=253ohm.

The diode doesn't do you any good for now, simply get rid of it, double check the polarity of the 24V signal and the pin orientation of the opto-isolator. With a resistor of anything between 1 and 2.7k you should read a voltage of around 1.2V across pin 1 and 2 of the PC817, if you still se 24V then the PC817 is shot.

/Henrik.

EDIT: Short answer to the question if it can replace a 78nn voltage regulator is no. You can use a resistor to "drop" voltage like is being done above but the current needs to be known and constant or the "output" voltage will go up and down as the current changes.

[Sunmixed]

Originally Posted by H.O Hi, If you have 24V across pin 1 and pin 2 of the PC817 then you either have the pins reversed or the LED in the opto-isolator is shot. LEDs are current driven and the have a nominal forward voltage drop. In the case of the PC817 this voltage drop is 1.2V typical and the forward current should be somewhere between 10 and 50mA depending on how much current you need to pull thru the transistor on the other side. If you want to drive the PC817 with a 24V signal, then 22.8V needs to be dropped across the series resistor. Let's say you want 15mA thru the LED then Ohms law (R=U/I) tells you the resistor needs to be 22.8V/0.015=1520ohm. If you want to drive the same PC817 with a 5V signal then resistor needs to be 3.8/0.015=253ohm. THe diode doesn't do you any good for now, simply ger rid of it, double check the polarity of the 24V signal and the pin orientation of the opto-isolator. With a resistor of anything between 1 and 2.7k you should read a voltage of around 1.2V across pin 1 and 2 of the PC817, if you still se 24V then the PC817 is shot. /Henrik.

Henrik, I did try your method earlier before posting. For some what reason, it still doesnt work. Like you said, in between 1 and 2.7k, my original resistor was 2.4K. Still, it ended up at 23.98. But i'm not quite sure with the current. Wouldn't you think that that little resistor is too big for 2AMPs? Should I get something higher for your proposed method?

While measuring the voltage, i've removed the PC817. Guess I will try it again, but with the optocoupler mounted.

Sunmixed

[H.O]

Hi,
It doesn't matter if the powersupply can deliver 0.2A, 2A or 200A in this case - you only need 0.015A and as long as it can deliver that amount it's fine.

The circuit will only pull the amount of current from the supply that it "needs" no matter how many amps the supply can deliver. Once again Ohm's law dictates, Current=Voltage/Resistance. So if you put your 2.2k resistor between + and - on the supply there will be a current of 24/2200=0.011A or 11mA flowing even if you have a 24V, 200A supply.

Do you have an ordinary LED handy? If so, replace the PC817 with the LED then you can see if lights up or not. Make sure you get the polarity right.

If you remove the PC817 there's definetly no current flowing and when there's no current flowing thru the resistor there won't be any voltage dropped across it so you'll get 24V on both sides of it. See, it's Ohm's law again, Voltage=Current*Resistance. Zero current=zero voltage drop results in the same voltage on both sides.

/Henrik.

[Zig]

All good suggestions.

I would suggest that you further try connecting an LED in series with the Optocoupler LED to check if it is in fact conducting current.